(1). I don't know what you mean by perfect, but it is correct. ${\mathbb Z}_4$ has an element of order 4, and ${\mathbb Z}_2^3$ hasn't, so no subgroup of ${\mathbb Z}_2^3$ is isomorphic to ${\mathbb Z}_4$.
(2). ${\mathbb Z}_2^3$ has 8 elements. One of them is the neutral element; all seven others have order 2.
(3). This doesn't have to be immediately clear at this point; the explanation is in the lines below. Up to isomorphism, the only groups of order 4 are ${\mathbb Z}_4$ and ${\mathbb Z}_2^2$, and in view of (1), subgroups of ${\mathbb Z}_2^3$ are not isomorphic to ${\mathbb Z}_4$. Hence the Klein group is mentioned because, whatever subgroup of order 4 you're going to find, it's going to be isomorphic to the Klein group (and not ${\mathbb Z}_4$ as just explained).
(4). This is true in general for an abelian group. The elements $x$ and $y$ are of order 2 and are distinct. Because $x$ is of order 2, $-x = x$, so $y \neq -x$, so $x + y \neq 0$. It does hold that $(x + y) + (x + y) = x + x + y + y = 0$, so $x + y$ has order 2.
(5). $\{i,x,y,x+y\}$ has 4 elements. What is left to show is that it is a subgroup, i.e., that it is closed under the group operations. For inversion: that's easy, as in ${\mathbb Z}_2^3$ every element is its own inverse anyway. For closure under addition: there are two remaining cases $x + (x + y)$ and $y + (x + y)$ and they are treated by $(\dagger)$ and $(\ddagger)$.
(6). The reasoning above says that among the 21 ways of writing down $i, x, y, x+y$ (with $x, y$ distinct and not $i$; and $x,y$ considered the same as $y,x$), there are 7 distinct sets $\{i, x, y, x + y\}$. The seven concrete distinct ways come from just starting to listing them and throwing out duplicates. Start with $\{i, a, b, a+b\}$, then $\{i, a, c, a+c\}$. After that $\{i, a, d, a+d\}$ turns out to be the same as $\{i, a, b, a+b\}$ as $a + d = b$, etc.
Edit: naming all the elements is maybe not the easiest way to list the subgroups of order 4, but is probably the most elementary way to find them. Using the symmetry inherent in ${\mathbb Z}_2^3$ the subgroups of order 4 can be described as follows. The subgroups are ${\mathbb Z}_2 \times {\mathbb Z}_2 \times \{0\}$ (plus two more of this form); the subgroup $\langle (1,1) \rangle \times {\mathbb Z}_2$ (plus two more of this form); the final one is the one generated by $(1,1,0)$ and $(0,1,1)$. They can also be seen as lines in the Fano plane. Anyway, for someone doing this particular exercise, simply listing them seems the best way to go about this.
(7). Right before (5) it is proven that they are subgroups.
This answer addresses the question in the title, not the many all along the post.
The elements of order $2$ of $G:=\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_4$ are:
\begin{alignat}{1}
&a_1=(0,0,2), \space a_2=(0,1,2), \space a_3=(1,0,2), \space a_4=(1,1,2) \\
&a_5=(0,1,0), \space a_6=(1,0,0), \space a_7=(1,1,0) \\
\tag 1
\end{alignat}
Any candidate subgroup of $G$, say $K$, isomorphic to the Klein group must be made of:
- the unit $1_G:=(0,0,0)$;
- any pair $a_i, a_j$;
- the element $a_i+a_j$; in fact: $a_i+a_j\in K$ by closure, and $a_i+a_j\ne 1,a_i,a_j$.
Is, for every $1\le i<j\le 7$, the subset $K_{ij}:=\{1_G,a_i,a_j,a_i+a_j\}$ indeed a subgroup of $G$? It's enough to prove the closure:
- $a_i^2=a_j^2=(a_i+a_j)^2=1_G$
- $a_i+(a_i+a_j)=(a_i+a_i)+a_j=a_j$
- $a_j+(a_i+a_j)=a_j+(a_j+a_i)=(a_j+a_j)+a_i=a_i$
So, indeed $K_{ij}\le G$ and $K_{ij}\cong\mathbb{Z}_2\times\mathbb{Z}_2$.
Now, if we denote $a_k:=a_i+a_j$, then $a_k+a_i=a_j$ and $a_k+a_j=a_i$. So:
- if $k<i<j$, then $K_{ij}=K_{ki}=K_{kj}$;
- if $i<k<j$, then $K_{ij}=K_{ik}=K_{kj}$;
- if $i<j<k$, then $K_{ij}=K_{ik}=K_{jk}$.
Therefore, the number of (distinct) subgroups of $G$ isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$ is:
\begin{alignat}{1}
n_K &= \frac{1}{3}\cdot|\{K_{ij}, \space1\le i<j\le 7\}| \\
&= \frac{1}{3}\cdot\frac{7\cdot 7-7}{2} \\
&= 7 \\
\tag 2
\end{alignat}
Explicitly, according to the labelling $(1)$:
\begin{alignat}{1}
K_{12} &= \{1_G,a_1,a_2,a_5\}\space (=K_{15}=K_{25}) \\
K_{13} &= \{1_G,a_1,a_3,a_6\}\space (=K_{16}=K_{36}) \\
K_{14} &= \{1_G,a_1,a_4,a_7\}\space (=K_{17}=K_{47}) \\
K_{23} &= \{1_G,a_2,a_3,a_7\}\space (=K_{27}=K_{37}) \\
K_{24} &= \{1_G,a_2,a_4,a_6\}\space (=K_{26}=K_{46}) \\
K_{34} &= \{1_G,a_3,a_4,a_5\}\space (=K_{35}=K_{45}) \\
K_{56} &= \{1_G,a_5,a_6,a_7\}\space (=K_{57}=K_{67}) \\
\tag 3
\end{alignat}
Best Answer
Suppose this non-cyclic subgroup is generated by elements (a,0), (b,1) (in other cases it would be cyclic, also any non-cyclic subgroup of a two-generated abelian group will be two-generated) and $a$ is nonzero. If $a$ generates $\mathbb{Z}_4$, then $\langle(a, 0), (b, 1)\rangle = \mathbb{Z}_4 \times \mathbb{Z}_2$. If $b$ generates $\mathbb{Z}_4$ and $a$ does not, then $\langle(a, 0), (b, 1)\rangle$ is cyclic. All other situations can be listed further: $\langle(2,0), (0, 1)\rangle$ and $\langle(2,0), (0, 1)\rangle$ both generate the same subgroup isomorphic to Klein four-group (which appears to be the only non-cyclic proper subgroup)